This document provides information about finding bases and dimensions of subspaces related to matrices. It defines the row space and column space of a matrix, and explains that the row rank and column rank of a matrix are equal. The dimension of the null space gives the number of free variables, while the column rank equals the number of pivot columns. Examples show how to find bases for the row space, column space, and null space of a matrix by row reducing to echelon form.

1. Find a Basis and Dimension of a Subspace
Find a basis and the dimension of the subspace
3 6
5 4
: , , , in
2
5
a b c
a d
H a b c d
b c d
d
   
      
   
    

2. Find a Basis and Dimension of a Subspace
H is the set of all linear combinations of the vectors
1
1
5
v
0
0
 
 
 
 
 
 
2
3
0
v
1
0
 
 
 
 
 
 
3
6
0
v
2
0
 
 
 
 
 
 
4
0
4
v
1
5
 
 
 
 
 
 
v3 = -2 v2
v1, v2, v4 are linearly independent

3. Find a Basis and Dimension of a Subspace
Therefore, {v1, v2, v4} is a set of linearly
independent vectors.
Also, H = span {v1, v2, v4}.
Hence {v1, v2, v4} is a basis for H.
Thus dim H =3.

4. 4. Vector Spaces
ROW and COLUMN
SPACES of a MATRIX
and
RANK of a MATRIX

5.  Defn - Let A be an m x n matrix. The rows of A,
considered as vectors in Rn , span a subspace of Rn
called the row space of A.
Each row has n entries, so Row A is a subspace of Rn.
11 12 1
21 22 2
1 2
n
n
mnm m
a a a
a a a
a a a
 
 
 
 
 
  
A
Row Space of a Matrix : RowA

6.  Defn - Let A be an m x n matrix. The columns of
A, considered as vectors in Rm, span a subspace of
Rm called the column space of A
 The system Ax = b has a solution if and only if b
is in the column space of A.
 Each column has m entries, so Col A is a subspace
of Rm.
11 12 1
21 22 2
1 2
n
n
mnm m
a a a
a a a
a a a
 
 
 
 
 
  
A
Column Space of a Matrix : Col A

7.  Theorem - If A and B are mxn row (column)
equivalent matrices, then the row (column) spaces
of A and B are equal (the same).
 Recall: If A and B are row equivalent when the
rows of B are obtained from the rows of A by a
finite number of the three elementary row
operations.
Equality of Spaces

8.  Defn - The dimension of the row space of a matrix
A is called the row rank of A
 Defn - The dimension of the column space of a
matrix A is called the column rank of A
 Theorem - The row rank and the column rank of
an mxn matrix A = [ aij ] are equal.
Rank of A

9.  The rank of A corresponds to the number of linearly
independent rows (columns) of A.
 If A is an n x n matrix, then rank A = n if and only if A is
row equivalent to In
 Let A be an n x n matrix. A is nonsingular if and only if
rank A = n
 If A is an n x n matrix, then rank A = n if and only if
det(A) ≠ 0.
 The homogeneous system Ax = 0 where A is an n x n
matrix has a nontrivial solution if and only if rank A < n
 Let A be an n x n matrix. The linear system Ax = b has a
unique solution if and only if rank A = n.
Rank of A

10. Equivalent statements for nxn matrix A
 A is nonsingular
 AX  0 has the trivial solution only
 A is row equivalent to In
 The system AX  b has a unique solution
 A has rank n
 The rows (columns) of A form a linearly independent
set of vectors in Rn
Rank of A

11. Let A be an mxn matrix.
Then
rank A + nullity A = n.
Nullity A = dim Null(A)
Rank + Nullity Theorem

12.  Example:
a. If A is a matrix with a two-dimensional
null space, what is the rank of A?
b. Could a matrix have a two-dimensional
null space?
 Solution:
a. Since A has 9 columns, , and
hence rank .
b. No. If a matrix, call it B, has a two-
dimensional null space, it would have to have
rank 7, by the Rank Theorem.
7 9
6 9
(rank ) 2 9A  
7A 
6 9
Rank + Nullity Theorem

13. To identify the basis for Row A and Col A
1. Reduce A to echelon form E
2. The nonzero rows of E form a basis for Row A
3. The columns of A that correspond to the columns of E
containing the leading entries form a basis for Col A
Note: The same is true if the matrix E is in reduced
echelon form.
For a matrix A, the number of linearly independent rows
is equal to the number of linearly independent columns.
Bases for Row A and Col A

14. has m = n = 2 and rank r = 1
Column space contains all multiples of
Row space consists of all multiples of [ 1 2 ]
Null space consists of all multiples of
1 2
3 6
 
 
 
A
1
3
 
 
 
1
2
 
 
 
Example

15. Example - Find a basis for the subspace V of R3
spanned by the rows of the matrix A
Row Space of a Matrix : RowA
1 2 1
1 1 1
1 3 3
3 5 1
1 4 5
 
 
 
 
 
 
 
 


 


A

16.  Example (continued) - Apply elementary row
operations to A to reduce it to B, in echelon form
1 0 3
0 1 2
0 0 0
0 0 0
0 0 0
 
 
 
 
 
 
 
 


B
A basis for the row space of B
consists of the vectors
w1 = [ 1 0 3 ] and w2 = [ 0 1 2 ].
So { w1, w2 } is a basis for V
v1 = w1  2w2, v2 = w1 + w2,
v3 = w1  3w2, v4 = 3w1  5w2,
v5 = w1  4w2
Row Space of a Matrix : RowA

17. Example (continued) - Since a matrix A is row
equivalent to a matrix B that is in row echelon form,
then
the nonzero rows of B form a basis for the row space
of A and the row space of B.
Row Space of a Matrix : RowA
w1 = [ 1 0 3 ] and w2 = [ 0 1 2 ].

18. DIMENSIONS OF NUL A AND COL A
 Let A be an matrix, and suppose the equation
has k free variables.
 A spanning set for Nul A will produce exactly k
linearly independent vectors—say, —one for
each free variable.
 So is a basis for Nul A, and the number of
free variables determines the size of the basis.
m n
x 0A 
1
u ,...,uk
1
{u ,...,u }k

19. DIMENSIONS OF NUL A AND COL A
 The dimension of Nul A is the number of free
variables in the equation , and the dimension
of Col A is the number of pivot columns in A.
 Example: Find the dimensions of the null space and
the column space of
x 0A 
3 6 1 1 7
1 2 2 3 1
2 4 5 8 4
A
   
   
 
   

20. DIMENSIONS OF NUL A AND COL A
 Solution: Row reduce the matrix to echelon form:
 There are three free variable—x2, x4 and x5.
 Hence the dimension of Nul A is 3.
 Also dim Col A = 2 because A has two pivot
columns (column 1 of A and column 3 of A).
1 2 2 3 1
0 0 1 2 2
0 0 0 0 0
  
 
 
  

21. Find bases for the row space, the column space, and the
null space of the matrix
2 5 8 0 17
1 3 5 1 5
3 11 19 7 1
1 7 13 5 3
   
 
 
 
   
Example

22. To find bases for the row space and the column space,
row reduce A to an echelon form:
 The first three rows of B form a basis for the row
space of A (as well as for the row space of B).
 Basis for Row A:
Example (cont) : Row space
1 3 5 1 5
0 1 2 2 7
0 0 0 4 20
0 0 0 0 0
A B
 
  
 
 
 
 
{(1,3, 5,1,5),(0,1, 2,2, 7),(0,0,0, 4,20)}   

23.  For the column space, observe from B that the pivots
are in columns 1, 2, and 4.
 Hence columns 1, 2, and 4 of A (not B) form a basis
for Col A:
Basis for Col
 Notice that any echelon form of A provides (in its
nonzero rows) a basis for Row A and also identifies
the pivot columns of A for Col A.
Example (cont) : Column space
2 5 0
1 3 1
: , ,
3 11 7
1 7 5
A
       
      
       
      
            

24.  However, for Nul A, we need the reduced echelon
form.
 Further row operations on B yield
Example (cont) : Null space
1 0 1 0 1
0 1 2 0 3
0 0 0 1 5
0 0 0 0 0
A B C
 
 
 
 
 
 

25.  The equation is equivalent to , that is,
 So , , , with x3
and x5 free variables.
x 0A  x 0C 
1 3 5
2 3 5
4 5
0
2 3 0
5 0
x x x
x x x
x x
  
  
 
1 3 5
x x x   2 3 5
2 3x x x  4 5
5x x
Example (cont) : Null space

26.  The calculations show that
Basis for Nul A
 Observe that, unlike the basis for Col A, the bases for
Row A and Nul A have no simple connection with the
entries in A itself.
Example (cont) : Null space
1 1
2 3
: ,1 0
0 5
0 1
     
         
    
    
    
        

27. Section 4.5
p.268, p.269
1 to 16
Problems